Example 9.12: Empirical/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas

Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?

Solution

\(f_{\mathrm{C}}\) \(= 0.857\)


\(f_{\mathrm{H}}\) \(= 0.143\)


\(m_{\mathrm{gas}}\) \(= 1.56\ \mathrm{g}\)


\(V_{\mathrm{gas}}\) \(= 1.00\ \mathrm{L}\)


\(P\) \(= 0.984\ \mathrm{atm}\)


\(T\) \(= 50\ \mathrm{°aC}\)


Strategy 1: Figure out the mass of carbon and hydrogen from mass and mass fractions, and the respective chemical amounts via the molar mass of these elements. Then, figure out the chemical amount of the gas molecule via the ideal gas law. The ratio of amount of atom vs amount of molecule gives the coefficient for that type of atom.

\(m_{\mathrm{C}}\) \(= f_{\mathrm{C}} \cdot m_{\mathrm{gas}}\)

\(\ \ \ =0.857 \cdot 1.56\ \mathrm{g}\)

\(\ \ \ =1.337\ \mathrm{g}\)


\(m_{\mathrm{H}}\) \(= f_{\mathrm{H}} \cdot m_{\mathrm{gas}}\)

\(\ \ \ =0.143 \cdot 1.56\ \mathrm{g}\)

\(\ \ \ =0.223\ \mathrm{g}\)


\(n_{\mathrm{C}}\) \(= \dfrac{m_{\mathrm{C}}}{M_{\mathrm{\ce{C}}}}\)

\(\ \ \ =\dfrac{1.337\ \mathrm{g}}{12.011\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =0.1113\ \mathrm{mol}\)


\(n_{\mathrm{H}}\) \(= \dfrac{m_{\mathrm{H}}}{M_{\mathrm{\ce{H}}}}\)

\(\ \ \ =\dfrac{0.223\ \mathrm{g}}{1.008\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =0.221\ \mathrm{mol}\)


Now we use the ideal gas law to figure out the chemical amount of the gas molecule:

\(P \cdot V_{\mathrm{gas}} = n_{\mathrm{molecule}} \cdot R \cdot T\)     

\(R\) \(= 8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)


\(n_{\mathrm{molecule}}\) \(= \dfrac{P \cdot V_{\mathrm{gas}}}{R \cdot T}\)

\(\ \ \ =\dfrac{0.984\ \mathrm{atm} \cdot 1.00\ \mathrm{L}}{8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} \cdot \frac{6463 }{ 20}\ \mathrm{K}}\)

\(\ \ \ =\dfrac{0.984\ \mathrm{L}\ \mathrm{atm}}{2686.67\ \frac{\mathrm{J}}{\mathrm{mol}}}\)

\(\ \ \ =0.0371\ \mathrm{mol}\)


\(\mathrm{coeff}_{\mathrm{C}}\) \(= \dfrac{n_{\mathrm{C}}}{n_{\mathrm{molecule}}}\)

\(\ \ \ =\dfrac{0.1113\ \mathrm{mol}}{0.0371\ \mathrm{mol}}\)

\(\ \ \ =3.00\)


\(\mathrm{coeff}_{\mathrm{H}}\) \(= \dfrac{n_{\mathrm{H}}}{n_{\mathrm{molecule}}}\)

\(\ \ \ =\dfrac{0.221\ \mathrm{mol}}{0.0371\ \mathrm{mol}}\)

\(\ \ \ =5.96\)


Molecular formula is \(\ce{C3H6}\)


Strategy 2: First solve the empirical formula problem using methods discussed earlier. Assume 1.56 g (because we actually know the mass of the sample). After calculating the chemical amounts of atoms as in strategy 1, divide by the smallest value (amount of carbon) to get the stoichiometric ratio \(n_{\mathrm{\ce{C}}}\) : \(n_{\mathrm{\ce{H}}}\). In the last step, realize that the smallest whole number ratio is the empirical formula:

\(\mathrm{ratio}_{\mathrm{\ce{H : C}}}\) \(= \dfrac{n_{\mathrm{H}}}{n_{\mathrm{C}}}\)

\(\ \ \ =\dfrac{0.221\ \mathrm{mol}}{0.1113\ \mathrm{mol}}\)

\(\ \ \ =1.99\)


The ratio of H to C is very close to 2, so the empirical formula is \(\ce{CH2}\), and we can calculate the molar mass of this empirical formula:

\(M_{\mathrm{\ce{CH2}}}\) \(= M_{\mathrm{\ce{CH2}}}\)

\(\ \ \ =14.027\ \frac{\mathrm{g}}{\mathrm{mol}}\)


Then, we calculate the molar mass of the gas molecule by dividing the mass of the sample by the chemical amount (obtained from ideal gas law):

\(M_{\mathrm{molecule}}\) \(= \dfrac{m_{\mathrm{gas}}}{n_{\mathrm{molecule}}}\)

\(\ \ \ =\dfrac{1.56\ \mathrm{g}}{0.0371\ \mathrm{mol}}\)

\(\ \ \ =42.0\ \frac{\mathrm{g}}{\mathrm{mol}}\)


Because the molar mass of the molecule is about three time higher than that of the empirical formula, we have to multiply the subscripts in the empirical formula \(\ce{CH2}\) by three to obtain the molecular formula, \(\ce{C3H6}\).